Question

A steel rod of length 2l, cross sectional area A and mass M is set rotating in a horizontal plane about an axis passing through the centre. If Y is the Young’s modulus for steel, find the extension in the length of the rod. (Assume the rod is uniform.)

Answer 1

Consider an element at r of width dr. Let (r) and T (r + dr) be the tension at the two edges, respectively.

Net centrifugal force in element = ω²rμdr

[∵ μ = mass/length]

−T(r + dr) + T(r) = μw²rdr

− \(\frac{dT}{dr}dr\) = μw²rdr

−dT = μw²rdr

∵ Tension and centrifugal force are opposite.

∴ \(∫^T_{T=0}dt\) = \(∫^{r=r}_{r=l}μ\text w^2rdr\)

T = \(μ\text w^2\frac{r^2}{2}|^l_r+C\)

T = \(\frac{μ ω^2}{2}\)(l² - r²) + C

At r = l, T = 0

T(r) = \(\frac{μ \text w^2}{2}\) (l² - r²)

Let the increase in length of the element dr be dδ therefore, Young’s Modulus,

Y = \(\frac{stress}{strain}\)

= \(\frac{\frac{T(r)}{A}}{\frac{dδ}{dr}}\)

Y = \(\frac{\frac{(\frac{μ\text w^2}{2})(l^2-r^2)}{A}}{\frac{d(δ)}{dr}}\)

∴ \(\frac{dδ}{dr}\) = \(\frac{1}{YA}\)\(\frac{μ\text w^2}{2}\)(l² - r²)

∴ dδ = \(\frac{1}{YA}\) \(\frac{μ\text w^2}{2}\)(l² - r²)dr

Change in length, δ = \(\frac{1}{YA}\)\(\frac{μ\text w^2}{2}\)\(∫^1_0\)(l² - r²)dr

= \(\frac{1}{YA}\)\(\frac{μ\text w^2}{2}\)[l³ − \(\frac{l^3}{3}\)] = \(\frac{1}{3YA}\)μw²l³

The total change in length is 2δ = \(\frac{2}{3YA}\)μw²l³